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My question is the following: if we had the trajectory of a particle eventually reaching a point of a rotation axis $ \vec{u} $ (take that as being the z-axis for convenience) by an angle $ s $, would Noethers Theorem still give a conserved quantity?

More specifically (let me go through the calculations and details first)

  1. Statement of Noether's Theorem

If a Lagrangian $ \mathcal{L}(\vec{q_i}, \dot{\vec{q_i}}, t) $ admits a one-parameter group of diffeomorphisms $ h^s : \mathcal{M} \rightarrow \mathcal{M} $ such that $ h^{(s=0)} (\vec{q_i})= \vec{q_i} $, then there is a conserved quantity locally given by $$I = \sum_{i} \frac{\partial \mathcal{L}}{\partial \dot{q_i}} \left.\frac{d}{ds}(h^s(q_i))\right\vert_{s=0}$$

  1. Applying to Simple Lagrangian

Assume a potential-free Lagrangian $ \mathcal{L} = \frac{m}{2}( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 ) $. A suitable transformation can be given by $$ h^s (x,y,z)= \begin{pmatrix} \cos(s) & -\sin(s) & 0 \\ \sin(s) & \cos(s) & 0 \\ 0 & 0 & 1 \end{pmatrix} $$

Working out the conserved quantity, we get that the z-component of angular momentum $ L_z = m \dot{y}(t) x(t) - m \dot{x}(t) y(t) $ is conserved for any path $ (x(t),y(t),z(t)) $.

  1. The problem: If this trajectory would include any point on the rotation axis z, $ h^s(q_i) $ would be 0 there and so by conservation, valued 0 all along the path. However, we know that angular momentum is conserved. So, in all rigor - is this inconsistency amendable or a sign of some bigger problem?
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2 Answers 2

There is no inconsistency. $L_z$ is conserved, but $h^s$ is not; hence it is not a paradox that $h^s$ has no action on the trajectory at a certain time and nontrivial action at others.

On the other hand, if the trajectory ever does cross the $z$ axis, then $L_z=m(\dot y x-\dot x y)$ will also vanish. Since $L_z$ is conserved, you can conclude that $L_z\equiv0$ for all time. This means that the motion of the particle will be constrained to the plane spanned by the $z$ axis and the particle's velocity when it crosses the $z$ axis.

To prove this, you can see the equation $L_z=0$ as a differential equation, which you can phrase as $$\frac{\dot y} y=\frac{\dot x}x,$$ and which integrates easily to $$\log(y)=\int\frac{\text dy}{y}=\int\frac{\text dx}{x}=\log(x)+C$$ so $x$ and $y$ are proportional, and therefore constrained to a plane. This is what you can conclude from the particle ever crossing the $z$ axis in a system with rotational symmetry about that axis.

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OP wrote (v3):

Working out the conserved quantity, we get that the $z$-component of angular momentum $ L_z = m x(t)\dot{y}(t) - m y(t)\dot{x}(t)$ is conserved for any path $(x(t),y(t),z(t))$.

Well, it should be stressed that the conservation law in Noether's theorem is an so-called on-shell conservation law. It does not hold for any curved (off-shell) path that one could cook up. It only holds for classical paths, i.e. solutions to Euler-Lagrange equations. For OP's example of a free particle this means constant velocity along a straight line. If such a classical trajectory intersects the $z$-axis (at one point), then the particle would indeed have angular momentum $L_z(t)=0$ for all $t$.

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